1. Advanced Algorithms Piyush Kumar (Lecture 2: Max Flows) Welcome to COT5405 Slides based on Kevin Wayne’s slides

2. Announcements Preliminary Homework 1 is out Scribing is worth 5% extra credit. My compgeom.com web pages might not be accessible from inside campus(for now you can use cs.fsu.edu/~piyush pages.

3. Today More about the programming assignment. On Max Flow Algorithms

4. Soviet Rail Network, 1955 Reference: On the history of the transportation and maximum flow problems. Alexander Schrijver in Math Programming, 91: 3, 2002. “The Soviet rail system also roused the interest of the Americans, and again it inspired fundamental research in optimization.” -- SchrijverSchrijver G. Danzig* 1951…First soln… Again formulted by Harris in 1955 for the US Airforce (“unclassified in 1999”) What were they looking for?

5. Maximum Flow and Minimum Cut Max flow and min cut. –Two very rich algorithmic problems. –Cornerstone problems in combinatorial optimization. –Beautiful mathematical duality.

6. Applications –Network reliability. –Distributed computing. –Egalitarian stable matching. –Security of statistical data. –Network intrusion detection. –Multi-camera scene reconstruction. –Many many more... - Nontrivial applications / reductions. - Data mining. - Open-pit mining. - Project selection. - Airline scheduling. - Bipartite matching. - Baseball elimination. - Image segmentation. - Network connectivity.

7. Some more history

8. Flow network. –Abstraction for material flowing through the edges. –G = (V, E) = directed graph, no parallel edges. –Two distinguished nodes: s = source, t = sink. –c(e) = capacity of edge e. Minimum Cut Problem s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 capacity source sink

9. Def. An s-t cut is a partition (A, B) of V with s  A and t  B. Def. The capacity of a cut (A, B) is : Cuts s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 Capacity = 10 + 5 + 15 = 30 A

10. s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 A Def. An s-t cut is a partition (A, B) of V with s  A and t  B. Def. The capacity of a cut (A, B) is: Cuts Capacity = 9 + 15 + 8 + 30 = 62

11. Min s-t cut problem. Find an s-t cut of minimum capacity. Minimum Cut Problem s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 A Capacity = 10 + 8 + 10 = 28

12. Def. An s-t flow is a function that satisfies: –For each e  E: (capacity) –For each v  V – {s, t}: (conservation) Def. The value of a flow f is: Flows 4 0 0 0 0 0 0 4 4 0 0 0 Value = 4 0 capacity flow s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 4

13. Flows as Linear Programs Maximize: Subject to:

14. Flows 10 6 6 11 1 10 3 8 8 0 0 0 11 capacity flow s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 Value = 24 4 Def. An s-t flow is a function that satisfies: –For each e  E: (capacity) –For each v  V – {s, t}: (conservation) Def. The value of a flow f is:

15. Max flow problem. Find s-t flow of maximum value. Maximum Flow Problem 10 9 9 14 4 10 4 8 9 1 0 0 0 14 capacity flow s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 Value = 28

16. Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s. Flows and Cuts 10 6 6 11 1 10 3 8 8 0 0 0 11 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 Value = 24 4 A

17. Flow value lemma. Let f be any flow, and let (A, B) be any s- t cut. Then, the net flow sent across the cut is equal to the amount leaving s. Flows and Cuts 10 6 6 1 3 8 8 0 0 0 11 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 Value = 6 + 0 + 8 - 1 + 11 = 24 4 11 A

18. Flow value lemma. Let f be any flow, and let (A, B) be any s- t cut. Then, the net flow sent across the cut is equal to the amount leaving s. Flows and Cuts 10 6 6 11 1 10 3 8 8 0 0 0 11 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 Value = 10 - 4 + 8 - 0 + 10 = 24 4 A

19. Flows and Cuts Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then Pf. by flow conservation, all terms except v = s are 0

20. Flows and Cuts Weak duality. Let f be any flow, and let (A, B) be any s-t cut. Then the value of the flow is at most the capacity of the cut. Cut capacity = 30  Flow value  30 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 Capacity = 30 A

21. Weak duality. Let f be any flow. Then, for any s-t cut (A, B) we have v(f)  cap(A, B). Pf. Flows and Cuts s t A B 7 6 8 4

22. Certificate of Optimality Corollary. Let f be any flow, and let (A, B) be any cut. If v(f) = cap(A, B), then f is a max flow and (A, B) is a min cut. Value of flow = 28 Cut capacity = 28  Flow value  28 10 9 9 14 4 10 4 8 9 1 0 0 0 14 s 2 3 4 5 6 7 t 15 5 30 15 10 8 15 9 6 10 15 4 4 0 A

23. Towards a Max Flow Algorithm Greedy algorithm. –Start with f(e) = 0 for all edge e  E. –Find an s-t path P where each edge has f(e) < c(e). –Augment flow along path P. –Repeat until you get stuck. s 1 2 t 10 00 00 0 20 30 Flow value = 0

24. Towards a Max Flow Algorithm Greedy algorithm. –Start with f(e) = 0 for all edge e  E. –Find an s-t path P where each edge has f(e) < c(e). –Augment flow along path P. –Repeat until you get stuck. s 1 2 t 20 Flow value = 20 10 20 30 00 00 0 X X X 20

25. Towards a Max Flow Algorithm Greedy algorithm. –Start with f(e) = 0 for all edge e  E. –Find an s-t path P where each edge has f(e) < c(e). –Augment flow along path P. –Repeat until you get stuck. greedy = 20 s 1 2 t 2010 20 30 200 0 opt = 30 s 1 2 t 2010 20 30 2010 20 locally optimality  global optimality

26. Residual Graph Original edge: e = (u, v)  E. –Flow f(e), capacity c(e). Residual edge. –"Undo" flow sent. –e = (u, v) and e R = (v, u). –Residual capacity: Residual graph: G f = (V, E f ). –Residual edges with positive residual capacity. –E f = {e : f(e) 0}. uv 17 6 capacity uv 11 residual capacity 6 flow

27. DemoDemo

28. Augmenting Path Algorithm Augment(f, c, P) { b  bottleneck(P) foreach e  P { if (e  E) f(e)  f(e) + b else f(e R )  f(e) - b } return f } Ford-Fulkerson(G, s, t, c) { foreach e  E f(e)  0 G f  residual graph while (there exists augmenting path P) { f  Augment(f, c, P) update G f } return f } forward edge reverse edge

29. Max-Flow Min-Cut Theorem Augmenting path theorem. Flow f is a max flow iff there are no augmenting paths. Max-flow min-cut theorem. [Ford-Fulkerson 1956] The value of the max flow is equal to the value of the min cut. Proof strategy. We prove both simultaneously by showing the TFAE: (i)There exists a cut (A, B) such that v(f) = cap(A, B). (ii)Flow f is a max flow. (iii)There is no augmenting path relative to f. (i)  (ii) This was the corollary to weak duality lemma. (ii)  (iii) We show contrapositive. –Let f be a flow. If there exists an augmenting path, then we can improve f by sending flow along path.

30. Proof of Max-Flow Min-Cut Theorem (iii)  (i) –Let f be a flow with no augmenting paths. –Let A be set of vertices reachable from s in residual graph. –By definition of A, s  A. –By definition of f, t  A. original network s t A B

31. Running Time Assumption. All capacities are integers between 1 and U. The cut containing s and rest of the nodes has C capacity. Invariant. Every flow value f(e) and every residual capacities c f (e) remains an integer throughout the algorithm. Theorem. The algorithm terminates in at most v(f*)  C iterations. Pf. Each augmentation increase value by at least 1. ▪ Corollary. If C = 1, Ford-Fulkerson runs in O(m) time. Integrality theorem. If all capacities are integers, then there exists a max flow f for which every flow value f(e) is an integer. Pf. Since algorithm terminates, theorem follows from invariant. ▪ Total running time?

32. 32 7.3 Choosing Good Augmenting Paths

33. Ford-Fulkerson: Exponential Number of Augmentations Q. Is generic Ford-Fulkerson algorithm polynomial in input size? A. No. If max capacity is C, then algorithm can take C iterations. s 1 2 t C C 00 00 0 C C 1 s 1 2 t C C 1 00 00 0 X 1 C C X X X 1 1 1 X X 1 1 X X X 1 0 1 m, n, and log C

34. Choosing Good Augmenting Paths Use care when selecting augmenting paths. –Some choices lead to exponential algorithms. –Clever choices lead to polynomial algorithms. –If capacities are irrational, algorithm not guaranteed to terminate! Goal: choose augmenting paths so that: –Can find augmenting paths efficiently. –Few iterations. Choose augmenting paths with: [Edmonds-Karp 1972, Dinitz 1970] –Max bottleneck capacity. –Sufficiently large bottleneck capacity. –Fewest number of edges.

35. Capacity Scaling Intuition. Choosing path with highest bottleneck capacity increases flow by max possible amount. –Don't worry about finding exact highest bottleneck path. –Maintain scaling parameter . –Let G f (  ) be the subgraph of the residual graph consisting of only arcs with capacity at least . 110 s 4 2 t 1 170 102 122 GfGf 110 s 4 2 t 170 102 122 G f (100)

36. Capacity Scaling Scaling-Max-Flow(G, s, t, c) { foreach e  E f(e)  0   smallest power of 2 greater than or equal to max c G f  residual graph while (   1) { G f (  )   -residual graph while (there exists augmenting path P in G f (  )) { f  augment(f, c, P) update G f (  ) }    / 2 } return f }

37. Capacity Scaling: Correctness Assumption. All edge capacities are integers between 1 and C. Integrality invariant. All flow and residual capacity values are integral. Correctness. If the algorithm terminates, then f is a max flow. Pf. –By integrality invariant, when  = 1  G f (  ) = G f. –Upon termination of  = 1 phase, there are no augmenting paths. ▪

38. Capacity Scaling: Running Time Lemma 1. The outer while loop repeats 1 +  log 2 C  times. Pf. Initially  < C.  drops by a factor of 2 each iteration and never gets below 1. Lemma 2. Let f be the flow at the end of a  -scaling phase. Then the value of the maximum flow is at most v(f) + m . Lemma 3. There are at most 2m augmentations per scaling phase. –Let f be the flow at the end of the previous scaling phase. –L2  v(f*)  v(f) + m (2  ). –Each augmentation in a  -phase increases v(f) by at least . Theorem. The scaling max-flow algorithm finds a max flow in O(m log C) augmentations. It can be implemented to run in O(m 2 log C) time. ▪ proof on next slide

39. Capacity Scaling: Running Time Lemma 2. Let f be the flow at the end of a  -scaling phase. Then value of the maximum flow is at most v(f) + m . Pf. (almost identical to proof of max-flow min-cut theorem) –We show that at the end of a  -phase, there exists a cut (A, B) such that cap(A, B)  v(f) + m . –Choose A to be the set of nodes reachable from s in G f (  ). –By definition of A, s  A. –By definition of f, t  A. original network s t A B

40. Bipartite matching. Can solve via reduction to max flow. Flow. During Ford-Fulkerson, all capacities and flows are 0/1. Flow corresponds to edges in a matching M. Residual graph G M simplifies to: –If (x, y)  M, then (x, y) is in G M. –If (x, y)  M, the (y, x) is in G M. Augmenting path simplifies to: –Edge from s to an unmatched node x  X. –Alternating sequence of unmatched and matched edges. –Edge from unmatched node y  Y to t. Bipartite Matching st 1 1 1 YX

41. References R.K. Ahuja, T.L. Magnanti, and J.B. Orlin. Network Flows. Prentice Hall, 1993. (Reserved in Dirac) R.K. Ahuja and J.B. Orlin. A fast and simple algorithm for the maximum flow problem. Operation Research, 37:748- 759, 1989. K. Mehlhorn and S. Naeher. The LEDA Platform for Combinatorial and Geometric Computing. Cambridge University Press, 1999. 1018 pages.The LEDA Platform for Combinatorial and Geometric Computing On the history of the transportation and maximum flow problems, Alexander Schrijver.On the history of the transportation and maximum flow problems